Food at College
# From your file menu (at the top of R-Studio) select:
# "Session -> Set working directory -> To source file location"
# Then play this chunk to get the data into R.
library(mosaic)
library(car)
library(DT)
library(pander)
library(readr)
library(tidyverse)
library(plotly)
food <- read_csv("food.csv") #food.csv is in the Data folder...food1 <- food |>
mutate(exercise = case_when(
exercise == 1 ~ "Everyday",
exercise == 2 ~ "2~3 Times a Week",
exercise == 3 ~ "Once a Week")) |>
mutate(GPA = as.numeric(GPA), exercise = as.factor(exercise)) |>
na.omit(GPA)Intro/Context
In the data set “Food” information about health habits, GPA, and basic demographic can be found. For this analysis I will only use the GPA that each person reported and how often they exercise in a week. There are three different groups 1.Everyday exercise, 2.Twice or Three Times a Week (2~3 Times a Week), and 3.Once a Week. I will measure if there is an impact on the GAP based on how often people exercise in a week regarding of their gender, height, and other information that will not be used for this analysis. The level of significance for this test will be α = 0.05.
The null hypotheses claims that all three groups have the same average GPA, represented by μ. The alternative hypotheses claims that at least one of the groups has a significantly different average GPA.
Ho/Ha (ANOVA)
\[ H_0: \mu \text {Everyday} = \mu \text{Twice or Three Times a Week} = \mu \text{Once a Week} = \mu \]
\[ H_a: \mu_i \neq \mu \] \[ α=0.05 \]
Graphical summary
ggplotly(ggplot(food1, aes(x=exercise, y=GPA)) +
geom_boxplot(fill=c("orange", "seagreen", "royalblue"), color="black")+
labs(title="Distribution of GPA by Group", x="Exercise Acitivity", y="GPA"))It is interesting to see this graph because it is commonly believed that daily exercise helps with students to perform better at school, but as we can see here the median for people who only exercise 2 or 3 times a week is slightly higher than people who exercise everyday. In fact, the median GPA for people who exercise every day is closer to the one for people who only exercise one. It might be because they take more time working out than studying or it could be that they put more effort at working out than at school. With an anova test we will be able to see if the average GAP is the same for all three groups or if at least one of them has a significantly different result.
Numerical summary
favstats(GPA ~ exercise, data = food1) |>
pander()| exercise | min | Q1 | median | Q3 | max | mean | sd | n | missing |
|---|---|---|---|---|---|---|---|---|---|
| 2~3 Times a Week | 3 | 3.3 | 3.65 | 3.8 | 4 | 3.565 | 0.3156 | 16 | 0 |
| Everyday | 2.8 | 3.2 | 3.5 | 3.7 | 3.904 | 3.446 | 0.3313 | 31 | 0 |
| Once a Week | 2.6 | 3.4 | 3.4 | 3.4 | 3.7 | 3.3 | 0.4123 | 5 | 0 |
The group with the highest GAP report was the 2~3 Times a Week with a max score of 4.0, and the Everyday exercise group reported a max score of 3.904. Does that mean that Everyday exercise people are giving up 0.096 of their GAP score for working out everyday? Not necessarily, we need to do an ANOVA test to keep everything into account and see if the average differs for at least one group or if they all perform equally based on average and not only the maximum score reported.
ANOVA test
food.aov <- aov(GPA ~ exercise, data = food1)
pander(summary(food.aov))| Df | Sum Sq | Mean Sq | F value | Pr(>F) | |
|---|---|---|---|---|---|
| exercise | 2 | 0.3066 | 0.1533 | 1.374 | 0.2627 |
| Residuals | 49 | 5.466 | 0.1116 | NA | NA |
The P-value for this is test is 0.2627 which is greater than the significance level used for this test of α = 0.05.
Check the requirements for the ANOVA tests
par(mfrow=c(1,2))
plot(food.aov, which= 1:2)
These two graph show the Residual vs Fitted on the left and the Normal Q-Q plot on the right. We can see that there are some outliers as shown in both graphs, and it’s becuase of these inconsistencies that a non-parametric test like Kruskal-Wallis would be more appropiate for this analysis.
Kruskal Wallis - compare results with ANOVA
kruskal.test(GPA ~ exercise, data=food1) |>
pander()| Test statistic | df | P value |
|---|---|---|
| 2.513 | 2 | 0.2847 |
The P-value of the ANOVA test was p-value = 0.2627, and for the Kruskal was p-value = 0.2847. The p-value for the Kruskal test was higher for 0.022. The degrees of freedom were the same for each test, df = 2. Even though, the p-value was higher for the Kruskal test, both ANOVA and Kruskal p-values are higher than the significance level of 0.05, which means that the both will fail to reject the null hypotheses.
Conclusion/Future Studies
The ANOVA test had a p-value = 0.2627 > α = 0.05. We conclude that we fail to reject the null hypotheses, accepting that overall the three different groups have the no significantly difference on average of GPA score.
food1 |>
group_by(exercise) |>
summarise(GPA_Average = mean(GPA)) |>
pander()| exercise | GPA_Average |
|---|---|
| 2~3 Times a Week | 3.565 |
| Everyday | 3.446 |
| Once a Week | 3.3 |
For future studies, the sample size of each group should be greater than 30. In that way we can fulfill the requirements of the anova tests and have a more consistent and large data set to analyze in more depth the hypotheses used for this analysis.